| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding minimum n for P(X=0) threshold |
| Difficulty | Standard +0.3 Part (i) is a standard Poisson approximation to binomial with straightforward calculation of P(X>3). Part (ii) requires solving P(X=0) = e^(-λ) < 0.01 for n, involving logarithms but following a routine method. Both parts are typical S2 applications with no novel insight required, making this slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Po(4) | M1, A1 | Use of Poisson, any mean; Correct mean |
| \(1 - e^{-4}(1 + 4 + \frac{4^2}{2!} + \frac{4^3}{3!})\) | M1 | Allow one end error |
| \(= 1 - 0.43347\ldots = 0.567\) or \(0.566\) | A1[4] | SC1: \(\frac{3.5 \cdot 4}{\sqrt{3.9984}}\) B1; SC2: Correct Bin method M1 ans 0.567 or 0.566 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = \frac{n}{2500}\) | B1 | \(\left(\frac{2499}{2500}\right)^n\) |
| \(e^{-\frac{n}{2500}} < 0.01\) | M1 | \(\left(\frac{2499}{2500}\right)^n < 0.01\); Correct exp'n \(< 0.01\). Allow '\(=\)' |
| \(-\frac{n}{2500} < \ln 0.01\) | \(n \times \ln\left(\frac{2499}{2500}\right) < \ln 0.01\) | |
| \(n > 11512.9\ldots\) Smallest \(n = 11513\) | A1[3] | \(n > 11510.6\ldots\) Smallest \(n=11511\); Allow by trial |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Po(4) | M1, A1 | Use of Poisson, any mean; Correct mean |
| $1 - e^{-4}(1 + 4 + \frac{4^2}{2!} + \frac{4^3}{3!})$ | M1 | Allow one end error |
| $= 1 - 0.43347\ldots = 0.567$ or $0.566$ | A1[4] | SC1: $\frac{3.5 \cdot 4}{\sqrt{3.9984}}$ B1; SC2: Correct Bin method M1 ans 0.567 or 0.566 A1 |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = \frac{n}{2500}$ | B1 | $\left(\frac{2499}{2500}\right)^n$ |
| $e^{-\frac{n}{2500}} < 0.01$ | M1 | $\left(\frac{2499}{2500}\right)^n < 0.01$; Correct exp'n $< 0.01$. Allow '$=$' |
| $-\frac{n}{2500} < \ln 0.01$ | | $n \times \ln\left(\frac{2499}{2500}\right) < \ln 0.01$ |
| $n > 11512.9\ldots$ Smallest $n = 11513$ | A1[3] | $n > 11510.6\ldots$ Smallest $n=11511$; Allow by trial |
---4 On average, 1 in 2500 people have a particular gene.\\
(i) Use a suitable approximation to find the probability that, in a random sample of 10000 people, more than 3 people have this gene.\\
(ii) The probability that, in a random sample of $n$ people, none of them has the gene is less than 0.01 . Find the smallest possible value of $n$.
\hfill \mbox{\textit{CAIE S2 2011 Q4 [7]}}